Cryptography - An art and science

Cryptography: An Art and Science

Sujit Prakash Gujar

sujit@csa.iisc.ernet.inIndian Institute of Science, Bangalore-12

September 15, 2007.

Sujit Prakash Gujar (CSA, IISc) Cryptography: An Art and Science September 15, 2007. 1 / 22

Agenda

Motivation

Objectives of Cryptography.Evolution: Need for Mathematical tools in cryptography.Encryption/Decryption: Art and Science.Hard Problems.RSA: Integer Factorization.Conclusions.

Agenda

MotivationObjectives of Cryptography.

Evolution: Need for Mathematical tools in cryptography.Encryption/Decryption: Art and Science.Hard Problems.RSA: Integer Factorization.Conclusions.

Agenda

MotivationObjectives of Cryptography.Evolution: Need for Mathematical tools in cryptography.

Encryption/Decryption: Art and Science.Hard Problems.RSA: Integer Factorization.Conclusions.

Agenda

MotivationObjectives of Cryptography.Evolution: Need for Mathematical tools in cryptography.Encryption/Decryption: Art and Science.

Hard Problems.RSA: Integer Factorization.Conclusions.

Agenda

MotivationObjectives of Cryptography.Evolution: Need for Mathematical tools in cryptography.Encryption/Decryption: Art and Science.Hard Problems.

RSA: Integer Factorization.Conclusions.

Agenda

MotivationObjectives of Cryptography.Evolution: Need for Mathematical tools in cryptography.Encryption/Decryption: Art and Science.Hard Problems.RSA: Integer Factorization.

Conclusions.

Agenda

MotivationObjectives of Cryptography.Evolution: Need for Mathematical tools in cryptography.Encryption/Decryption: Art and Science.Hard Problems.RSA: Integer Factorization.Conclusions.

Motivation

Colonel wants to convey some military plan of action to LieutenantColonel.

World War I: Zimmermann Telegram.World War II:Famous for cipher breaking.On-line banking.

Motivation

Colonel wants to convey some military plan of action to LieutenantColonel.World War I: Zimmermann Telegram.

World War II:Famous for cipher breaking.On-line banking.

Motivation

Colonel wants to convey some military plan of action to LieutenantColonel.World War I: Zimmermann Telegram.World War II:Famous for cipher breaking.

On-line banking.

Motivation

Colonel wants to convey some military plan of action to LieutenantColonel.World War I: Zimmermann Telegram.World War II:Famous for cipher breaking.On-line banking.

Objectives of Cryptography

Privacy: CiphersAuthentication: Digital Signatures.Data Integrity: Message DigestNon-repudiation: Digital Signatures.

Evolution of Cryptography

Steganography: art and science of writing hidden messages.

Demeratus: Wooden Tablet. (440 BC)Substitution Ciphers.Statistical Analysis.

Steganography: art and science of writing hidden messages.Demeratus: Wooden Tablet. (440 BC)

Substitution Ciphers.Statistical Analysis.

Steganography: art and science of writing hidden messages.Demeratus: Wooden Tablet. (440 BC)Substitution Ciphers.

Statistical Analysis.

Steganography: art and science of writing hidden messages.Demeratus: Wooden Tablet. (440 BC)Substitution Ciphers.Statistical Analysis.

Evolution of CryptographyPolyalphabetic substitution

Vigenere cipher: polyalphabetic substitution.

Enigma.

Vigenere cipher: polyalphabetic substitution.Enigma.

Use of Keys.Secret Algorithms vs. Secret Key Algorithms.Distribution of keys?

Cipher: Art and Science

Cipher: an encryption and decryption algorithm.Encryption: Scrambles message data based on key. Mathematically,

E : M ×K → C

E (m, ke) = c

Decryption: recovers message data,only when provided correct key.Mathematically,

D : C ×K → M

D(c, kd) = m

Desirable properties:Easy to encrypt.Difficult to invert without key.

Some Standard Ciphers

Simple way, c = m ⊕ k, m′ = c ⊕ k = m ⊕ k ⊕ k = m.

Symmetric key cryptosystems, i.e. ke = kd

– Block ciphers: AES,DES,IDEA– Stream ciphers: RC4,FISH,SEAL

Public key cryptosystems, i.e. ke 6= kdRSA,Rabbin,ECC. (Elliptic Curve Cryptosystem)Need of Mathematical Tools, i.e functions which are easy to computeand difficult to invert.

Simple way, c = m ⊕ k, m′ = c ⊕ k = m ⊕ k ⊕ k = m.Symmetric key cryptosystems, i.e. ke = kd

– Block ciphers: AES,DES,IDEA

– Stream ciphers: RC4,FISH,SEAL

Public key cryptosystems, i.e. ke 6= kdRSA,Rabbin,ECC. (Elliptic Curve Cryptosystem)

Need of Mathematical Tools, i.e functions which are easy to computeand difficult to invert.

Preliminaries

We say, a ≡ b mod (n), when n divides (a− b).It is basically a remainder function.84 ≡ 75 mod (9), but in general, we will say 84 ≡ 3 mod (9)

When n = p, some prime, p,Let, Z×p = {1, 2, . . . , p − 1}.

(Z×p ,×p): Multiplicative group.

g ∈ Z×p such that, {g , g2, . . . , gp−1} = Z×

p , then we say g isgenerator.

Hard Problems

Integer Factorization Problem: (IFP)Given an integer n = p ∗ q, product of two prime numbers, find primefactors, p, q.

Discrete Logarithm Problem (DLP)Let p be the prime number and g be the generator of (Zp)×.If, a = g x mod (p), x is called as discrete logarithm of a.DLP is, given ‘a’ find discrete logarithm of ‘a’ in (Zp)×

i.e. find x s.t. g x mod (p) = a.Diffie-Hellman Problem: (DHP)Let p be the prime number and g be the generator of (Zp)×.Given ga mod (p),gb mod (p), find gab mod (p)

Hard Problems

Integer Factorization Problem: (IFP)Given an integer n = p ∗ q, product of two prime numbers, find primefactors, p, q.Discrete Logarithm Problem (DLP)Let p be the prime number and g be the generator of (Zp)×.If, a = g x mod (p), x is called as discrete logarithm of a.DLP is, given ‘a’ find discrete logarithm of ‘a’ in (Zp)×

i.e. find x s.t. g x mod (p) = a.

Diffie-Hellman Problem: (DHP)Let p be the prime number and g be the generator of (Zp)×.Given ga mod (p),gb mod (p), find gab mod (p)

Hard Problems

Integer Factorization Problem: (IFP)Given an integer n = p ∗ q, product of two prime numbers, find primefactors, p, q.Discrete Logarithm Problem (DLP)Let p be the prime number and g be the generator of (Zp)×.If, a = g x mod (p), x is called as discrete logarithm of a.DLP is, given ‘a’ find discrete logarithm of ‘a’ in (Zp)×

i.e. find x s.t. g x mod (p) = a.Diffie-Hellman Problem: (DHP)Let p be the prime number and g be the generator of (Zp)×.Given ga mod (p),gb mod (p), find gab mod (p)

RSA Cryptosystem consists Three Primitives:

Key Generation Encryption Decryption

Key Generation1 Choose, p and q, two large primes.2 Calculate: n = p ∗ q. φ(n) = (p − 1) ∗ (q − 1).3 Choose e such that g.c.d of e and φ(n) is 1.4 Calculate d such that ed ≡ 1 mod (φ(n)).

RSA: Key Generation

If g.c.d of a and b is 1, there exists unique x and y such that

ax + by = 1.

Let, a = e, b = φ(n). Then, x will be required d .And this can be calculated using Extended Euclidian Algorithm.(e, n) is called public key or encryption key.(d , n) is called private key or decryption key.

RSA Contd...

Let, (e, n) be public key of A and (d , n) corresponding private key.Message m, is such that 1 ≤ m ≤ n

EncryptionWhen B want to send a m to A,he will sendc ≡ me mod (n)

DecryptionAfter receiving c,A will retrieve message back usingm′ ≡ cdmod(n)

m′ ≡ cd

≡ (me)d mod (n)

≡ m(ed) mod (n)

When, n = p ∗ q, from Number Theory, we haveax = a(x+φ(n)−1) mod (n)m′ ≡ m mod (n).

Cracking RSA

Look at Key generation. If somebody solves IFP, he can easily deducedecryption key.Note: But, given public key, if someboy can deduce private keydoesn’t implay he can factor n. i.e.Solving IFP ⇒ Cracking RSAReverse may not be true.Rabin Cryptosystem: As hard as IFP.

Factoring number

Naive algorithm: Try all numbers 1 to√

n.Time complexity: O(

Suppose, x2 ≡ y2 mod n and x 6≡ ±y mod (n)⇒ n divides (x2 − y2) i.e. either (x + y) or (x − y).Who will give such x and y?

Factoring numberDixton,quadratic sieving,number field sieving

Choose B, set of known prime numbers.Choose randomly, x1, x2, . . . , xk , s.t. p(xi ) ≡ x2

i mod (n) factorscompletely using primes in B.Find {y1 . . . , yl} ⊆ {x1, x2, . . . , xk}, such that

l∏i=1

y2i ≡

l∏i=1

p(yi ) mod (n)

and, RHS of the above equation is square.Quadratic Sieving. Time Complexity: O

(exp(2

√2√

log n log log n)).

Number Field Sieving. Time Complexity:O

(exp(c ∗ (log n)1/3(log log n)2/3)

Elliptic curve cryptosystems

y2 = x3 + ax + b.Consider, set of all integer pair (x,y) s.t.

y2 = x3 + ax + b mod (p),

where p is prime.Group.We can use DLP.Till date, Time complexity: O

(exp(c(log n)1/2(log log n)1/2)

1024 bit RSA security ≡ 168 bit ECC security.

Conclusions

Need of Mathematical functions in cryptography.Hard Problems.RSA CryptosystemInteger Factoring.Elliptic Curve Cryptosystems.

References

“Applied Cryptography”, (2nd Ed.) by Bruce Schneier.“Handbook of Applied Cryptography”, by Alfred Menezes, Paul vanOorschot and Scott Vanstone.“The Code Book”, by Simon Singh.

Questions?

Thank You!!!

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